MANE 4240 & CIVL 4240 Introduction to Finite Elements

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MANE 4240 & CIVL 4240 Introduction to Finite Elements. Prof. Suvranu De. Introduction to 3D Elasticity. Reading assignment: Appendix C+ 6.1+ 9.1 + Lecture notes. Summary: 3D elasticity problem Governing differential equation + boundary conditions Strain-displacement relationship

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## MANE 4240 & CIVL 4240 Introduction to Finite Elements

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### MANE 4240 & CIVL 4240Introduction to Finite Elements

Prof. Suvranu De

Introduction to 3D Elasticity

Appendix C+ 6.1+ 9.1 + Lecture notes

• Summary:
• 3D elasticity problem
• Governing differential equation + boundary conditions
• Strain-displacement relationship
• Stress-strain relationship
• Special cases
• 2D (plane stress, plane strain)
• Principle of minimum potential energy

y

A(x) = cross section at x

b(x) = body force distribution (force per unit length)

E(x) = Young’s modulus

u(x) = displacement of the bar at x

F

x

x

x=L

x=0

1. Strong formulation: Equilibrium equation + boundary conditions

Equilibrium equation

Boundary conditions

3. Stress-strain (constitutive) relation :

E: Elastic (Young’s) modulus of bar

2. Strain-displacement relationship:

Surface (S)

Volume (V)

w

v

u

z

x

y

x

3D Elasticity

Problem definition

V: Volume of body

S: Total surface of the body

The deformation at point

x =[x,y,z]T

is given by the 3 components of its displacement

NOTE:u= u(x,y,z), i.e., each displacement component is a function of position

3D Elasticity:

EXTERNAL FORCES ACTING ON THE BODY

• Two basic types of external forces act on a body
• Body force (force per unit volume) e.g., weight, inertia, etc
• Surface traction (force per unit surface area) e.g., friction

Volume element dV

Xc dV

Xb dV

Xa dV

w

Volume (V)

v

u

Surface (S)

z

x

y

x

BODY FORCE

Body force: distributed force per unit volume (e.g., weight, inertia, etc)

NOTE: If the body is accelerating, then the inertia force

may be considered as part of X

Volume element dV

pz

Xc dV

py

Xb dV

px

Xa dV

w

Volume (V)

v

ST

u

z

x

y

x

SURFACE TRACTION

Traction: Distributed force per unit surface area

3D Elasticity:

INTERNAL FORCES

w

v

u

sz

Volume element dV

tzy

tzx

tyz

txz

sy

txy

Volume (V)

tyx

sx

z

x

y

If I take out a chunk of material from the body, I will see that,

due to the external forces applied to it, there are reaction

forces (e.g., due to the loads applied to a truss structure, internal

forces develop in each truss member). For the cube in the figure,

the internal reaction forces per unit area(red arrows) , on each

surface, may be decomposed into three orthogonal components.

x

sz

tzy

tzx

tyz

txz

sy

txy

z

tyx

sx

y

x

3D Elasticity

sx, sy and sz are normal stresses.

The rest 6 are the shear stresses

Convention

txy is the stress on the face perpendicular to the x-axis and points in the +ve y direction

Total of 9 stress components of which only 6 are independent since

The stress vector is therefore

Strains: 6 independent strain components

Consider the equilibrium of a differential volume element to obtain the 3 equilibrium equations of elasticity

Compactly;

EQUILIBRIUM EQUATIONS

(1)

where

3D elasticity problem is completely defined once we understand the following three concepts

• Strong formulation (governing differential equation + boundary conditions)
• Strain-displacement relationship
• Stress-strain relationship

Volume element dV

pz

Xc dV

py

Xb dV

px

Xa dV

w

Volume (V)

v

ST

u

z

x

Su

y

x

1. Strong formulation of the 3D elasticity problem: “Given the externally applied loads (on ST and in V) and the specified displacements (on Su) we want to solve for the resultant displacements, strains and stresses required to maintain equilibrium of the body.”

(1)

Equilibrium equations

Boundary conditions

1. Displacement boundary conditions: Displacements are specified on portion Su of the boundary

2. Traction (force) boundary conditions:Tractions are specified on portion ST of the boundary

Now, how do I express this mathematically?

Volume element dV

pz

Xc dV

py

Xb dV

px

Xa dV

w

Volume (V)

v

ST

u

z

x

Su

y

x

Traction: Distributed force per unit area

TS

nz

pz

n

py

ny

px

nx

ST

Traction: Distributed force per unit area

If the unit outward normal to ST :

Then

py

TS

txy

q

sx

ds

dy

px

y

txy

dx

x

sy

In 2D

n

ny

q

q

dy

nx

ds

dx

ST

Consider the equilibrium of the wedge in x-direction

Similarly

3D elasticity problem is completely defined once we understand the following three concepts

• Strong formulation (governing differential equation + boundary conditions)
• Strain-displacement relationship
• Stress-strain relationship

(2)

Compactly;

C’

y

In 2D

C

B’

A’

u

dy

v

A

B

dx

x

3D elasticity problem is completely defined once we understand the following three concepts

• Strong formulation (governing differential equation + boundary conditions)
• Strain-displacement relationship
• Stress-strain relationship

3. Stress-Strain relationship:

Linear elastic material (Hooke’s Law)

(3)

Linear elastic isotropic material

Special cases:

1. 1D elastic bar (only 1 component of the stress (stress) is nonzero. All other stress (strain) components are zero)

Recall the (1) equilibrium, (2) strain-displacement and (3) stress-strain laws

2. 2D elastic problems: 2 situations

PLANE STRESS

PLANE STRAIN

3. 3D elastic problem: special case-axisymmetric body with axisymmetric loading (we will skip this)

Area element dA

Nonzero stress components

h

D

Assumptions:

1. h<<D

2. Top and bottom surfaces are free from traction

3. Xc=0 and pz=0

y

x

PLANE STRESS Examples:

1. Thin plate with a hole

2. Thin cantilever plate

PLANE STRESS

Nonzero stresses:

Nonzero strains:

Isotropic linear elastic stress-strain law

Hence, the D matrix for the plane stress case is

y

x

z

PLANE STRAIN: Only the in-plane strain components are nonzero

Nonzero strain components

Area element dA

Assumptions:

1. Displacement components u,v functions of (x,y) only and w=0

2. Top and bottom surfaces are fixed

3. Xc=0

4. px and py do not vary with z

PLANE STRAIN Examples:

1. Dam

y

x

z

Slice of unit thickness

1

2. Long cylindrical pressure vessel subjected to internal/external pressure and constrained at the ends

PLANE STRAIN

Nonzero stress:

Nonzero strain components:

Isotropic linear elastic stress-strain law

Hence, the D matrix for the plane strain case is

y

2

1

2

2

4

3

x

Example problem

The square block is in plane strain and is subjected to the following strains

Compute the displacement field (i.e., displacement components u(x,y) and v(x,y)) within the block

Solution

Recall from definition

Arbitrary function of ‘x’

Integrating (1) and (2)

Arbitrary function of ‘y’

Plug expressions in (4) and (5) into equation (3)

Function of ‘y’

Function of ‘x’

Hence

Integrate to obtain

D1 and D2 are two constants of integration

Plug these back into equations (4) and (5)

How to find C, D1 and D2?

y

2

1

2

2

4

3

x

Use the 3 boundary conditions

To obtain

Hence the solution is

Principle of Minimum Potential Energy

Definition: For a linear elastic body subjected to body forces X=[Xa,Xb,Xc]T and surface tractions TS=[px,py,pz]T, causing displacements u=[u,v,w]T and strains e and stresses s, the potential energyP is defined as the strain energy minus the potential energy of the loads involving X and TS

P=U-W

Volume element dV

pz

Xc dV

py

Xb dV

px

Xa dV

w

Volume (V)

v

ST

u

z

x

Su

y

x

Strain energy of the elastic body

Using the stress-strain law

In 1D

In 2D plane stress and plane strain

Why?

Principle of minimum potential energy: Among all admissible displacement fields the one that satisfies the equilibrium equations also render the potential energy P a minimum.